Question

If (1 + i) (1 + 2i) (1 + 3i) ...... (1+ni) = a + ib, then 2.5.10.17 ...... ($1+n^2$) =

• A. a - ib
• B. $a^2-b^2$
• C. $a^2+b^2$
• D. none of these

Answer 1

The correct option is C

$a^2+b^2$

(1 + i) (1 + 2i) ( 1 + 3i)....(1 + ni) = a + ib

Taking modulus on both sides, we get

|(1+i) (1+2i) (1+3i) ..... (1+ni)| = a + ib

|(1 + i) (1 + 2i) (1 + 3i) .... (1 + ni)| can be writeen as |(1 + i)| |(1+2i)| |(1+3i)| ... |(1 + ni)|

$\therefore \sqrt{1^2+1^2}\times \sqrt{1^2+2^2}\times \sqrt{1^2+3^2}....\times \sqrt{1^2+n^2}=\sqrt{a^2+b^2}\Rightarrow \sqrt{2}\times \sqrt{5}\times \sqrt{10}.....\times \sqrt{1+n^2}=\sqrt{a^2+b^2}$

Squaring on both the sides, we get :

$2\times 5\times \mathrm{10.....}\times (1+n^2)=a^2+b^2$